Texas Waste Water Treatment Ch#15.2 Primary Clarifier

TLDR: Pi × R^2 × D × 7.48 gal/ft^3 = Gallons => 3.14 × 25 ft × 25 ft × 10 ft × 7.48 gal/ft^3 = 146,795 gal! ||| EXPLANATION: This question wants you to calculate the capacity in gallons of a cylindrical clarifier. To do this you first find the area of the circle (Area = Pi X R^2 = Pi X R X R = 3.14 X 25ft X 25ft = 1962.5 sqft.) then multiply by how deep the circle is to give you the volume of the cylinder. (Volume = Area X Depth = 1962.5 ft^2 X 10ft = 19625 cuft.) Now that we have the volume of the clarifier we need to calculate how many gallons of water will fit in that space. (Capacity = Volume X the number of gallons that fill 1 ft^3 of volume = 19625ft^3 X 7.48 gallons per ft^3 = 146,795 Gallons!) Pi in this instance will be limited to 3 significant figures. The diameter is halved to yeild the radius. (Radius = Diameter / 2 = 50ft / 2 = 25ft) Radius squared is simplified into multiplying the radius twice. (Radius^2 = R X R) Finally, the 7.48 gal/ft^3 is a conversion factor that should be provided on the reference sheet with the test but its a good number to know. Its this conversion factor that allows us to convert between volume and the gallons that fill that volume. In short the conversion factor is saying that in every cubic foot there are 7.48 gallons of water which allows us to say "if thats how much is in 1 cubic foot then we can multiply that number by the number of cubic feet that we have to get the total number of gallons that can fit in that volume".

TLDR: MG = gpd / 1,000,000 gal/MG = 170,000 gal / 1,000,000 gal/MG = 0.17 MG | lb = MGD× 8.34 lb/gal × mg/L = 0.17 × 8.34 lb/gal × 100 mg/L = 142 lb/day! ||| EXPLANATION: To start this I would like to mention that while the question as written in the book and copied here mentions BOD5 this was a measure meant to confuse the test taker. BOD technically stands for Biological Oxygen Demand which is the value the BOD5 test tells us after its 5 day run time. This means that BOD values given any time you see them would be 5 days old before we could perform any calculation. The delay does not meaningfully affect the determinations we use the values for such as unit or plant efficiency when you take into consideration the trending that is maintained with daily sampling. With that said, lets get into how we will answer this question. This is another problem that you can use the mnemonic in the top left corner for but there are some things that needs to be done first. For the the mnemonic to work it requires flow to be used in millions of gallons per day or it wont convert correctly because while 8.34 is the number of pounds that one gallon of water weighs it is ALSO a conversion of miligrams per liter to pounds per MILLION gallons. It is once again the conversion from mg/l to lbs/gal that we are using and so a value for flow in millions of gallons per day is required. Accordingly, we will start by converting the flow given in gallons per day to millions of gallons per day then set that number aside for later use. (MGD = GPD / 1,000,000 = 170,000GPD / 1,000,000 = 0.17 MGD) The next number we need to calculate, and it was calculated for us in the question, is how much BOD in mg/L we are removing in our primary clarifier. We do that by determining the difference between what comes in and what goes out. (mg/L removed = In - Out = 280mg/L - 180mg/L = 100mg/L) Now that we have both of those numbers we can use the mnemonic to answer the question. LBS/DAY is the value we are looking for, and the variables we have are flow in MGD, BOD removed in mg/L, and the conversion factor given in the mnemonic. The mnemonic tells us that the pounds of BOD removed per day is equal to the number of millions of gallons processed multiplied by the amount in lbs of BOD removed from 1 million gallons at a rate of 1mg/L (To restate "MGD X 8.34" tells us how much BOD in LBS would be removed if the BOD in each of those millions of gallons was 1mg/L) and then multiplied by the number of mg/L we are actually removing. (lbs of BOD removed per day = MGD X 8.34 lbs/million gallons X mg/L removed from the clarifier = 0.17MGD X 8.34 lbs/MG X 100mg/L = 142 lbs of BOD removed per day!)

TLDR: L × W × D × 7.48 gal/ft3 = Gallons 50 ft × 25 ft × 10 ft × 7.48 gal/ft3 = 93,500 gal | gph = gpd / 24 hrs/day = 1,500,000 gpd / 24 hr/day = 62,500 gph | DT in hrs = Capacity / Flow gph = 93,500 gal / 62,500 gph = 1.5 hrs DT! ||| EXPLANATION: To calculate detention time (the amount of time required for a unit of water to travel through a processing unit or in practice the amount of time required for the incoming flow to completely displace the volume in a processing unit) We first need to know the volume of the tank using the provided length, width, and depth. (volume = length X width X depth = 50ft X 25ft X 10ft = 12500ft^3) Then we take the volume and multiply that by the number of gallons that one cubic foot can hold to determine the capacity in gallons of the tank. (Capacity = Volume X 7.48 gal/ft^3 = 12500ft^3 X 7.48 gal/ft^3 = 93,500 gal) Then we need to convert our flow to a time scale that applies to the question asked. That just means that we divide our flow given in gallons per day by the number of hours in a day to determine the number of gallons recieved per hour. (gph = gpd / 24hrs per day(hpd) = 1,500,000gpd / 24hpd = 62,500gph) We now have everything we need to calculate detention time or how long it will take the flow coming into the tank to displace the volume already in the tank. We do this by dividing the capacity of the tank in gallons by the number of gallons per hour that will be displaced. (DT = Capacity / Flow = 93,500gal / 62,500gph = 1.5hrs!)

TLDR: Gallons = Pi × R^2 × D × 7.48 gal/ft3 = 3.14 × 30 ft × 30 ft × 10 ft × 7.48 gal/ft3 = 211,385 gal | gpd = MGD × 1,000,000 gal/MG = 1 MGD × 1,000,000 gal/MG = 1,000,000 gpd | gph = gpd / 24 hrs/day = 1,000,000 gpd / 24 hr/day = 41,667 gph | DT hrs = Capacity gal / Flow gph = 211,385 gal / 41,667 gph = 5 hrs! ||| EXPLANATION: Much like problem 3 we are calculating detention time in hours. Unlike last time its a cylindrical tank and our flow is in millions of gallons per day instead of gallons per day. I'm going to do it a little differently to show different methods and short cuts. Same as in problem three we need to calculate the capacity of the tank which is equal to the volume of the tank times our gallons per cubic foot conversion factor and set that number aside for later. (Cpacity = volume (of a cylinder) X gallons per cubic foot conversion factor = Pi X R^2 X D X gallons per cubic foot conversion factor = Pi X R X R X D X 7.48gal/ft^3 = Pi X (diameter / 2) X (diameter / 2) X D X 7.48gal/ft^3 = 3.14 X (60/2)ft X (60/2)ft X 10ft X 7.48gal/ft^3 = 3.14 X 30ft X 30ft X 10ft X 7.48gal/ft^3 = 211,385 gal) Next we shuold convert our flow given in millions of gallons per day to gallons per hour which is as easy as multiplying our MGD flow by one million to get that same value in GPD then dividing that number by the number of hours in a day. (gph = (MGD X 1,000,000gal/MG)gpd / 24 hrs/day) = (1MGD X 1,000,000gal/MG)gpd / 24 hrs/day = 1,000,000gpd / 24 hrs/day = 41,667gph) Now that we have capacity in gallons and flow in gph we can calculate how long it will take that flow to fill that tank and as a result displace the volume of fluid that was in there before. Detention time is equal to the volume in gallons divided by the number of gallons per hour being displaced. (DT = Capacity / Flow = 211,385gal / 41,667 gph = 5hrs!)

TLDR: MG/D/Clarifier = 1.0 MGD / 2 clarifiers = 0.5 MGD/clarifier | gpd = MGD × 1,000,000 gal/MG = 0.5 MG × 1,000,000 gal/MG = 500,000 gpd | Surface Area = Pi × R2 = 3.14 × 18ft × 18ft = 1017 ft2 | gal/ft^2/day = gal / ft^2 of surface area = 500,000 gpd / 1,017 ft2 = 492 gal/ft^2/day! ||| EXPLANATION: Surface loading, or overflow rate, is the most important consideration in the design of sedimentation basins. It is determined by dividing the quantity of wastewater flowing out of the tank by the surface area of the tank. Due to the fact we are dealing with an area instead of a volume the depth given is useless information, but the fact we are given combined flow for two clarifiers is not because we are asked to find individual (and identicle) surface loading values. To get started we need to divide our given total flow evenly between the two clarifiers so that we can perform our calculation for each clarifier with the correct flow value. (flow in each clarifier(MGD) = Millions of gallons of flow recieved per day / # of clarifiers = 1MGD / 2 clarifiers = 0.5MGD) Now that we have the correct flow we need to convert it to a format that matches the units we are about to be calculating meaning that we need to turn our MGD into GPD. (gpd = MGD X 1,000,000 = 0.5MGD X 1,000,000 = 500,000gpd) The next step now that we have our flow value in a format we can use is to calculate how many gallons pass over one square foot of tank per day. For this we will divide our flow by the calculated surface area of our tank based on the provided dimensions. (surface loading in gal/ft^2/day = flow in gpd / area in ft^2 = flow in gpd / (Pi X R^2)ft^2 = flow in gpd / (Pi X (dia. / 2)ft X (dia. / 2)ft)ft^2 = 500,000gpd / (3.14 X (36/2)ft X (36/2)ft)ft^2 = 500,000 gpd / (3.14 X 16ft X 16ft)ft^2 = 500,000 gpd / 1,017ft^2 = 492 gal/ft^2/day!)

TLDR: Diameter = 36 ft - 2 ft = 34 ft | linear weir length = Circumference = Pi × Diameter = 3.14 × 34 ft = 107 ft | gpd per clarifier = (MGD X 1,000,000)gpd / # of clarifiers = (4MGD X 1,000,000)gpd / 2 clarifiers = 4,000,000gpd / 2 clarifiers = 2,000,000gpd | gal/linear ft/day = gpd / linear ft of weir = 2,000,000gpd / 107ft = 18,692gal/linear ft/ day! ||| EXPLANATION: Weir loading also known as weir overflow rate is the number of gallons of water passing over a foot of weir per day, and we want to find that value based in part on information given in the previous problem. We still need to divide our flow equally between the two clarifiers and convert millions of gallons per day(MGD) to gallons per day(gpd) so lets get that out of the way and set that number aside for later. We need to take the peak flow it wants us to use multiply it by one million to convert from MGD to gpd then divide by the number of clarifiers given in the previous problem to determine the number of gallons transiting a single clarifier each day. (gpd per clarifier = (MGD X 1,000,000)gpd / # of clarifiers = (4MGD X 1,000,000)gpd / 2 clarifiers = 4,000,000gpd / 2 clarifiers = 2,000,000gpd) Next we need to know how many linear ft of weir we are dealing with. They tried to be sneaky and confuse you by telling us that the weir is set one foot inside the wall they gave us as part of the dimensions of the clarifier. This means that since the weir is ONE foot from the edge of the thirty six foot diameter on BOTH sides of the circle that it is really two feet we need to subtract from the given diameter. (I lack the ability to draw pictures or provide diagrams on these quizzes, but i invite you to examine and double check this assertion. To subtract only one foot from the diameter would describe a circle of weir that was one foot from the wall on one side and touches on the other or is only half a foot from the wall at all points. Weir diemeter = Wall diameter - 1ft at 0 degrees - 1ft at 180degrees = 36ft - 2ft = 34ft) Now we can divide the flow transiting the clarifier by the linear feet of weir we have to determine how many gallons are flowing over each linear foot of weir. (gal/linear ft/day = gpd / linear ft of weir = 2,000,000gpd / 107ft = 18,692gal/linear ft/day!)

TLDR: Settled = In - Out = 15 mL/L - 3 mL/L = 12 mL/L | Gallons to be removed = mL settled × 1,000 × MGD = 12 mL/L × 1,000 × 2.6 MGD = 31,200 gal! ||| EXPLANATION: The answer given in the answer key and copied to the TLDR section confused me a lot so im going to break it down as far as i can for you. The goal of this exercise is to figure out how much sludge should be pumped from a clarifier daily. The first step is to calculate how much sludge we are collecting. The test used to do this is the imhoff cone test which determines how much sludge is settling and staying in the clarifier by collecting a 1 leiter sample in an Imhoff cone from both the inlet and outlet ends of the clarifier. The samples then settle for 2 hours after which time the settled solids in each sample are measured based on the milliliter mark it settles to. This yeilds a unit that describes the number of milliliters of sludge obtained in the one leiter sample. we subtract the amount that settled from the effluent from the amount that settled from the influent to determine how much of the incoming solids remain in the clarifier to become sludge per leiter of waste water that passes through the clarifier. (milliliter settled per leiter of transiting waste water = mL/L of solids entering the clarifier - mL/L of solids leaving the clarifier = 15 mL/L sample - 3 mL/L sample = 12 mL/L transiting) Since we now know that for every leiter entering the clarifier we gain 12ml of sludge we could multiply this by the flow through the clarifier in leiters and have our answer but we are americans and like to make our lives complicated so this doesnt help us much. What the question wants to know is how many gallons of sludge we need to pump out daily so we start by converting our milileiters of sludge removed per leiter transiting (mL/L) that we got from the last step to leiters of sludge removed per leiter transiting (L/L). To make the milliliter unit cancel out, you divide by one thousand because thats how many millileiters there are in a leiter. (12 mL/L / 1000 ml/L = 0.012 leiters of sludge removed per leiter of waste water transiting the clarifier) so now we have leiters removed per leiter transiting and what we want is gallons removed per million gallons transiting. The next step, purely for illustration will convert L/L to gal/gal which is to say from leiters of sludge removed per leiter transiting to gallons removed per gallon transiting. You end up both multiplying and dividing by .264172 so the units and number cancel out leaving us with the same 0.012 we haad earlier but now its 0.012 gallons of sludge removed per gallon transiting. (gal/gal = (L removed per L transiting X gal per L conversion factor) / gal per L conversion factor = (0.012 L/L X .264172 gal/L) / .264172 gal/L = 0.012 gal/gal) Because our units still dont quite match up lets examine what values we have to work with now and what we are trying to accomplish. we now know that for every gallon transiting the clarifier we are removing 0.012 gallons of sludge that would otherwise just collect in the clarifier until it eventually over filled. we dont want that to happen so we need to find out how much sludge this means we are collecting in a day so that we can remove exactly that amount at a daily rate. The flow was given in the question in millions of gallons per day(MGD) that transit the clarifier but that doesnt quite match our current units of gallons removed per gallon transiting because its not in the same range so we will convert gallons to millions of gallons. This can be accomplished either in a second conversion focusing on MGD (gpd = MGD X 1,000,000) OR we can do it by changing the units on our removed sludge value (gallons of sludge removed per millions of gallons of flow(gal/MG) = gal/gal X gal/MG = 0.012 gal/gal X 1,000,000 gal/MG = 12,000 gallons of sludge removed per million gallons transiting the clarifier (gal/MG)) If, however, you combine the last few steps into one problem you end up with a conversion that divides by one thousand and multiplies by one million which is simplified in the textbook example by just multiplying by a unitless one thousand. (gal/MG = mL/L X 1000) either way you choose you now know how many gallons of sludge are removed per gallon (or million gallons) transiting the clarifier and flow in gallons per day or millions of gallons per day. All that remains is to multiply them together to find out how much sludge needs to be removed from the clarifier every day to offset how much sludge is being collected. (gallons to be removed from the clarifier each day(gpd) = gal/MG X MGD = gal/gal X gpd = 12,000gal/MG X 2.6MGD = 0.012gal/gal X 2,600,000gpd = 31,200gpd removed from clarifier!)